/*
Max Sum Plus Plus
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.
Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S 1, S 2, S 3 ... S n.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8

Hint
Huge input, scanf and dynamic programming is recommended.
题目大意：给定一个数组长度为m的数组，求其分成n个不相交子段和最大值的问题
*/
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
const int INF = 0x3f3f3f3f;
using namespace std;
int pre[1000010],num[1000010],n,m,a[1000010];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i = 1; i <= m; i ++)
            scanf("%d",&a[i]);
        memset(pre,0,sizeof(pre));
        memset(num,0,sizeof(num));
        int maxx;
        for(int i = 1; i <= n; i ++)
        {
            maxx = -INF;
            for(int j = i; j <= m; j ++)
            {
                num[j] = max(num[j - 1] + a[j],pre[j - 1] + a[j]);
                pre[j - 1] = maxx;
                if(num[j] > maxx)
                    maxx = num[j];
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}
